△ABC中,若cosA=3/5,sinB=5/13,则sinC=

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△ABC中,若cosA=3/5,sinB=5/13,则sinC=

△ABC中,若cosA=3/5,sinB=5/13,则sinC=
△ABC中,若cosA=3/5,sinB=5/13,则sinC=

△ABC中,若cosA=3/5,sinB=5/13,则sinC=

cosA=3/5
∵ 0∴ sinA>0
∵ sin²A=1-cos²A=1-9/25=16/25
∴ sinA=4/5
∵ sinB=5/13利用正弦定理
∴ b∴ B∴ B是锐角
∵ cos²B=1-sin²B=1-25/169=144/169
∴ cosB=12/13
∴ sinC
=sin[π-(A+B)]
=sin(A+B)
=sinAcosB+cosAsinB
=(4/5)*(12/13)+(3/5)*(5/13)
=48/65+15/65
=63/65

cosA=3/5,则:sinA=4/5,角A为锐角。
又:sinB=5/13
且:sinA>sinB,利用正弦定理得到:a>b,从而有:A>B
所以cosB=12/13
sinC=sin(A+B)=sinAcosB+cosAsinB=63/65

cosA=3/5,sinA=4/5
sinB=5/13,cosB=±12/13
sinC
=sin(A+B)
=sinAcosB+cosAsinB
=±4/5*12/13+3/5*5/13
=63/65或-33/65

cosA=3/5,sinB=5/13
sinA =4/5,cosB=±12/13 由(sina)^2+(cosa)^2=1,而 A、B、C均在0-180°间
sinC=sin(π-A-B)
=sin(A+B)
=(sinAcosB+cosAsinB)
=(4/5*±12/13+3/5*5/13)
=63 /65或者-33/65
故sinC=63/65

在△ABC中,已知cosA=3/5,sinB=5/13,则sinC=63/65
解:cosA=3/5→sinA=4/5
sinB=5/13→cosB=±12/13
(1).当cosA=3/5,sinA=4/5,sinB=5/13,cosB=12/13
sinC=sin[180°-(A+B)]=sin(A+B)=
sinAcosB+cosAsinB=...

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在△ABC中,已知cosA=3/5,sinB=5/13,则sinC=63/65
解:cosA=3/5→sinA=4/5
sinB=5/13→cosB=±12/13
(1).当cosA=3/5,sinA=4/5,sinB=5/13,cosB=12/13
sinC=sin[180°-(A+B)]=sin(A+B)=
sinAcosB+cosAsinB=(4/5)*(12/13)+(3/5)*(5/13)
=48/65+15/65=63/65
(2).当cosA=3/5,sinA=4/5,sinB=5/13,cosB=-12/13
sinC=sin[180°-(A+B)]=sin(A+B)=
sinAcosB+cosAsinB=(4/5)*(-12/13)+(3/5)*(5/13)
=-48/65+15/65=-33/65,但0°0
∴此种情况不合题意,舍去.
综上sinC=63/65

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